A rectangular sheet of metal of length 6m and width 2 m is given. Four equal squares are removed from the corners. The sides of this sheet are now turned up to form an open rectangular box. The height of box such that the volume of the box is maximum will be
Answers
The height of box is (4 - √7)/3 = 0.45 m such that the volume of the box is maximum
Step-by-step explanation:
A rectangular sheet of metal of length 6m and width 2 m is given
Four equal squares are removed from the corners
Let say Square size = x * x m
Length of open rectangular box = 6 - 2x m
width of open rectangular box = 2- 2x m
2 - 2x > 0
=> x < 1
Height = x m
Volume = (6 - 2x)(2 - 2x)x
= 2(3 - x)2(1 - x)x
=4x(3 + x² - 4x)
= 4x³ - 16x² + 12x
V = 4x³ - 16x² + 12x
dV/dx = 12x² - 32x + 12
12x² - 32x + 12 = 0
=> 3x² - 8x + 3 = 0
x = (8 ± √(64 - 36) )/2(3)
= (8 ± 2√7)/(2 * 3)
= ( 4 ± √7)/3
as x < 1
=> x = (4 - √7)/3 = 0.45
d²V/dx² = 24x - 32 is - ve as x < 1
Hence Volume is maximum
The height of box is (4 - √7)/3 = 0.45 m such that the volume of the box is maximum
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